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Suppose $f:X\to Y$ preserves $x_1\geq x_2\implies y_1\geq y_2$ but does not necessarily preserve $x_1> x_2\implies y_1> y_2$.

In other words, an $f$ satisfying this relation might, upon iteration, reduce a set to equality, or it might reduce any pair and everything in-between to equality, but cannot ever exchange the orders of any elements.

Does this have a name?

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  • $\begingroup$ do you mean $x_1\geq x_2\Rightarrow f(x_1)\geq f(x_2)$? $\endgroup$ – Pink Panther May 10 at 12:16
  • $\begingroup$ @PinkPanther yes. I'm hoping it's implicit $x_n\in X, y_n\in Y$ and $f(x_n)=y_n$. $\endgroup$ – user334732 May 10 at 12:18
  • $\begingroup$ That implication doesn't characterize what could be almost an isomorphism. For example, if $X$ is the four-element lattice that is not a chain and $Y$ is the four-element chain, and $f$ is a bijective map sending top to top and bottom to bottom, then $x_1 \leq x_2$ implies $f(x_1) \leq f(x_2)$, but the posets don't seem anything close to isomorphic. On the other hand notice that if the map satisfies that first implication but not the second, then it is not injective; again, far from isomorphism. $\endgroup$ – amrsa May 11 at 9:00
  • $\begingroup$ @amrsa it depends if "nearly" means not-order or not-isomorphic. On reflection, I mean specifically, not injective, and I think this is the correct interpretation of the answer given below. $\endgroup$ – user334732 May 11 at 9:04
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A function $f$ is order embedding when
$\forall x,y: (x \leq y \iff f(x) \leq f(y))$.

$f$ is an order isomorphism when $f$ is an order embedding bijection.

$f$ is order preserving when
$\forall x,y: (x \leq y \implies f(x) \leq f(y))$.

Exercise. If the domain of $f$ is linear and
$\forall x,y: (x < y \implies f(x) < f(y))$,
prove $f$ is order embedding.

The glibness of your question, including use of y for f(.) and lack of clarity about whether the order is linear or partial, leaves us confused about what you are asking. The use of y for f(.) should be discarded.

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    $\begingroup$ Remarks about the questsion clarity and glibness are more suited to comment imho. $\endgroup$ – Q the Platypus May 11 at 3:39

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